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\title{Introduction to Quantum Mechanics\\Lecture 14}
\author{Brian Greene\footnote{These lecture notes were TeX'd by Rohan Bhandari and Carlos Hernandez}} % ADD YOUR NAME HERE IF YOU WORK ON THIS LECTURE's NOTES
\date{November 1, 2011}

\maketitle

\section{Quantum Harmonic Oscillator}
This lecture focuses on solving the Time Independent Schrödinger Equation (TISE) for the
Quantum Harmonic Oscillator. This system is defined by the potential 
\begin{equation*}
V(x)=\frac{k {x}^{2}}{2} =\frac{m {w}^{2}{ x}^{2}}{2}
\end{equation*}
Our first step is now to
write out the TISE. Notice that if you factor the $\Psi$, you get the equation
\begin{equation*}
 (\frac{{p}^{2}}{2m} + \frac{m w {x}^{2}}{2})\Psi = E \Psi
\end{equation*} 
We can rewrite the left-hand side as an operator, $\hat{H}$, acting on $\Psi$.  $\hat{H}$ is an energy operator, when it operates on $\Psi$, you can retrieve $E$, the
energy of the system. We will from now on refer to this operator as the Hamiltonian.

\subsection{An Elegant Solution}
It is possible to directly solve this non-linear second order differential equation for the TISE, but it is a tough calculation. 
Instead, we will approach it in a much more elegant way.

Let's start with the formula $(A+B) =  (A+iB) (A-i B)$ to factor $\hat{H}$, into $a_+$  and $a_-$, where
\begin{equation}
a_+ = \frac{1}{\sqrt{2\hbar  m w}}(-i p + m w x)
\end{equation}
\begin{equation}
a_- = \frac{1}{\sqrt{2\hbar  m w}}(i p + m w x)
\end{equation}

Note, since we are dealing with operators, this formula is not entirely correct. Instead, we have
\begin{equation*}
a_+ a_- = \frac{1}{2 \hbar  m w} ({p}^{2} + {m}^{2} {x}^{2} {x}^{2} +i m w [x,p])
= \frac{1}{\hbar w}( (\frac{{p}^{2}}{2m} + \frac{m w}{2} {x}^{2}) - \frac{ \hbar  w}{2}) 
= \frac{1}{\hbar  w} (\hat{H} - \frac{\hbar  w}{2})
\end{equation*}
 This is because the momentum operator does not commute (i.e. [x,p]=  $x p - p x \neq 0$). This quantity is often referred to as the commutator of $x$ and $p$. Because of this descrepancy, we must
define $\hat{H}$ as
\begin{equation*}
\hat{H} = \hbar  w (a_+ a_- + \frac{1}{2})
\end{equation*}
Furthermore, we can now rewrite the TISE as
\begin{equation}   
E \Psi = \hbar  w (a_+ a_- + \frac{1}{2}) \Psi
\end{equation}

In addition, we'll make use of these identities later on:
\begin{equation} \label{eq:4}
[a_-,a_+] =  a_- a_+ - a_+ a_- = 1
\end{equation}
\begin{equation}
[\hat{H}, a_+] = \hat{H} a_+ -  a_+\hat{H} = \hbar  w  a_+
\end{equation}
\begin{equation}
[\hat{H}, a_-] = \hat{H} a_- -  a_-\hat{H} = \hbar  w  a_-
\end{equation}

\subsection{The Trimurti}
Now, suppose you are given $\Psi_{E}$ with energy $E$, yielding 
\begin{equation*}
\hat{H} \Psi = E \Psi
\end{equation*}
 Let us define a $\Psi_+$, such that $\Psi_+= a_+ \Psi_{E}$. Note that $\Psi_+$ is not necessarily normalized, even though $\Psi_{E}$ is. This then gives us
\begin{equation*}
\hat{H}\Psi_+ = \hat{H} a_+ \Psi_E = \hbar  w (a_+ a_- + \frac{1}{2})  a_+ \Psi_E = \hbar  w a_+ ( a_-  a_++ \frac{1}{2}) \Psi_E
\end{equation*}
Now, we can use Eqn.(\ref{eq:4}) to rewrite this as
\begin{equation*}
\hat{H}\Psi_+ = \hbar  w a_+ ( (a_+  a_- + \frac{1}{2}) + 1) \Psi_E = a_+ (\hat{H} + \hbar  w ) \Psi_E = (E + \hbar  w ) a_+ \Psi_E = E_+\Psi_+
\end{equation*}
Where $E +\hbar w$ is a constant, so we can break this statement up into two parts:
\begin{equation*}
E_+ = E + \hbar  w 
\end{equation*}
\begin{equation*}
\Psi_+ = a_+ \Psi_E
\end{equation*}
We now can see that the $a_+$ operator increases the energy of the state $\Psi_E$ by $\hbar w$. A similar derivation can be done with $a_-$:
\begin{equation*}
\hat{H}\Psi_- = \hat{H} a_- \Psi_E = \hbar  w (a_+ a_- + \frac{1}{2})  a_- \Psi_E
\end{equation*}
Again, we can apply Eqn.(\ref{eq:4}) to yield
\begin{equation*}
\hat{H}\Psi_- = \hbar  w  ( a_-  a_+ - \frac{1}{2}) a_- \Psi_E =  \hbar  w a_-  (( a_+ a_-   + \frac{1}{2}) - 1)  \Psi_E = a_-(\hat{H} - \hbar w)   \Psi_E = (E - \hbar w)  a_- \Psi_E E_- \Psi_-
\end{equation*}
Leaving us with
\begin{equation*}
E_- = E - \hbar w
\end{equation*}
\begin{equation*}
\Psi_- =   a_- \Psi_E
\end{equation*}
Where we can see that the operator $a_-$ decreses the energy of the state $\Psi_E$ by $\hbar w$. Note that for both of these operators, the resulting $\Psi_+$ or $\Psi_-$ is not necessarily normalized. Additionally, you can combine these operators as much as you'd like to yield a new wave state
\begin{equation*}
\Psi_{+++} =   a_+ a_+ a_+ \Psi_E
\end{equation*}
or even recover the same one
\begin{equation*}
\Psi_E = a_+ a_- \Psi_E
\end{equation*}
It is because of these properties that we refer to $a_+$ as the creation, or raising, operator and $a_-$ as the annihilation, or lowering, operator.
\subsection{$\Psi_0$}
As we've learned previously, the energy of a wavefunction cannot be less than the energy of the potential, $V$ acting on the particle. If it did violate this rule, there would be resulting negative energies, which do not have physical significance in the quantum theory we have developed thus far. Therefore, there must be a state, $\Psi_0$ that holds the lowest possible energy, $E_0$, for a system with a given potential. Using our knowledge of operators, we can define it as
\begin{equation} \label{eq:diff}
a_- \Psi_0 = 0
\end{equation}
We can this, in conjuction with our new definition of the Hamiltonian, to find the energy associated with this state in the case of the quantum harmonic oscillator
\begin{equation*}
\hat{H}\Psi_0 = \hbar  w (a_+ a_- + \frac{1}{2})\Psi_0 = \frac{\hbar  w}{2} \Psi_0 = E_0 \Psi_0 \to E_0 = \frac{\hbar  w}{2}
\end{equation*}
Since we now know there is a lower bound the allowable energy for a particle in this system, it follows that any state, $\Psi_k$, can be derived from acting on $\Psi_0$ with $k$ number of creation operators.
\begin{equation}
\Psi_{k} =   {a_+}^{k} \Psi_0
\end{equation}
\begin{equation}
E_k =   E_0 + k \hbar w  = \hbar w \frac{2k + 1}{2}
\end{equation}
\subsection{Uniqueness of the Solutions}
How do we know that the operators $a_+$ and $a_-$ give us the full set of solutions? Say there is another
wavefunction, $\phi_m$, that is a solution. Now if you apply $a_-$ on $\phi_m$ enough times, you get ${a_-}^{n+1} \phi_{m} = 0$,
which means that $\phi_{m-n} = \psi_0$ and, thus, $\phi_{m} = \psi_n$. Because there is minimum, non-negative energy, $E_0$, in the system, all solutions can be expressed as functions of $\psi_0$.
\subsection{Solving the TISE}
Let us now find the expression for $\psi_0$. Since Eqn.(\ref{eq:diff}) is essentially just a differential equation, we can use this to find a solution.
\begin{equation*} 
a_- \psi_0 = \frac{1}{\sqrt{2\hbar  m w}}(i p + m w x)\psi_0 = \frac{1}{\sqrt{2\hbar  m w}}(-\hbar\partial _x + m w x)\psi_0 = 0
\end{equation*}
\begin{equation*} 
\to  \hbar\partial_x \psi_0 = -m w x\psi_0 \to  \frac{\partial_x \psi_0}{\psi_0} = -\frac{m w x}{\hbar}
\end{equation*}
If we integrate both sides with respect to $x$ and normalize the result, we can retrieve
\begin{equation} 
\psi_0 = {(\frac{m w}{\pi \hbar})}^{\frac{1}{4}}{e}^{-\frac{m w }{2\hbar}{x}^{2}}
\end{equation}
Now that we have the solution to $\psi_0$, we can just use $a_+$ to solve for all possible solutions up to a factor of normalization. Let's demonstrate this for $\psi_2$
\begin{equation*} 
\psi_2 = A({ a_+}^{2}\psi_0) = \frac{A{(\frac{m w}{\pi \hbar})}^{\frac{1}{4}}}{2\hbar  m w}{(-i p + m w x)}^{2}{e}^{-\frac{m w }{2\hbar}{x}^{2}}
\end{equation*}
Let's simply this by not worrying about that nasty coefficient
\begin{equation*} 
\psi_2 = B {(-i p + m w x)}^{2}{e}^{-\frac{m w }{2\hbar}{x}^{2}} = B (-{p}^{2} +{(mwx)}^2 - i mwxp - i mwpx){e}^{-\frac{m w }{2\hbar}{x}^{2}}
\end{equation*}
Remember that $p$ is an operator ($p = -i\hbar\partial_x$), so those last two terms \emph{are} different
\begin{equation*} 
\psi_2 = B ({\hbar}^{2}{\partial_x}^{2} +{(mwx)}^2 - \hbar mwx\partial_x - mw\hbar){e}^{-\frac{m w }{2\hbar}{x}^{2}} 
\end{equation*}
\begin{equation*} 
= B(-\hbar m w{\partial_x}(x{e}^{-\frac{m w }{2\hbar}{x}^{2}}) + {(m w x)}^2 {e}^{-\frac{m w }{2\hbar}{x}^{2}} + {m}^{2}{w}^{2}x^2{e}^{-\frac{m w }{2\hbar}{x}^{2}} - \hbar m w {e}^{-\frac{m w }{2\hbar}{x}^{2}})
\end{equation*}
\begin{equation*} 
= B(-\hbar m w({e}^{-\frac{m w }{2\hbar}{x}^{2}} -\frac{m w }{\hbar} {x}^{2}{e}^{-\frac{m w }{2\hbar}{x}^{2}}) + {(m w x)}^2 {e}^{-\frac{m w }{2\hbar}{x}^{2}} + {m}^{2}{w}^{2}x^2 {e}^{-\frac{m w }{2\hbar}{x}^{2}} - \hbar m w {e}^{-\frac{m w }{2\hbar}{x}^{2}})
\end{equation*}
\begin{equation*}
=  m w B (2m w x^{2} -  h) e^{-\frac{m w x^2}{2 h}}
\end{equation*}
We can normalize this answer by demanding that 
\begin{equation*}
 \int _{-\infty }^{\infty } \psi_2^{*}\psi_2 dx = C^{2}\int _{-\infty }^{\infty }e^{-\frac{m w x^2}{h}} \left(- h+2 m w x^2\right)^2dx = 1
\end{equation*}
And find that $C = {(\frac{m w}{4\pi \hbar^{5}})}^{\frac{1}{4}}$. Therefore our final normalized solution for $\psi_2$ looks like
\begin{equation*}
\psi_2 =  {(\frac{m w}{4 \pi \hbar^5})}^{\frac{1}{4}}(2 m w x^{2} - h) e^{-\frac{m w x^2}{2 h}}
\end{equation*}
It turns out that in general, one can express a wavefunction as
\begin{equation}
\psi_n = \frac{1}{\sqrt{2^n n!}} {a_+}^n \psi_0
\end{equation}
We can use this to write the most general solution to for TISE
\begin{equation}
\Psi(x,t) = \sum_n \frac{C_n}{\sqrt{2^n n!}} {a_+}^n \psi_0(x) \phi_n(t)
\end{equation}
where
\begin{equation*}
C_n = \frac{1}{2L} \int _{-L }^{L }{\Psi}^*(x,0) \Psi(x,0) dx
\end{equation*}
\end{document}